Faults and Symmetrical Components
Symmetrical three-phase faults and fault MVA, sub-transient and transient reactances, symmetrical components, sequence networks, and line-to-ground, line-to-line and double line-to-ground faults.
A fault (short circuit) draws very large currents. Fault studies size circuit breakers and set protection.
Symmetrical (three-phase) faults
- With pre-fault voltage 1 pu and Thevenin impedance Z_th at the fault: I_f = 1/Z_th pu.
- Fault MVA = S_base/Z_th(pu) = √3 V_L I_f.
- A generator's reactance rises after the fault: sub-transient X″_d (first cycles) < transient X′_d < synchronous X_d (steady state). Breakers are rated using X″_d.
- Current-limiting reactors reduce fault levels.
Symmetrical components
Any unbalanced set of three phasors = sum of three balanced sets:
- Positive sequence (abc rotation), negative sequence (acb), zero sequence (all in phase).
- With operator a = 1∠120°: I_a0 = (I_a + I_b + I_c)/3, I_a1 = (I_a + a I_b + a² I_c)/3, I_a2 = (I_a + a² I_b + a I_c)/3.
- 1 + a + a² = 0.
- Zero-sequence current flows only if there is a return path (grounded neutral); the neutral carries 3I₀. Delta windings block zero sequence from the lines.
Unsymmetrical faults (unloaded generator, E = 1 pu)
| Fault | Connection of sequence networks | Fault current |
|---|---|---|
| Single line-to-ground (LG) | Series: Z₁ + Z₂ + Z₀ (+ 3Z_f) | I_f = 3E/(Z₁ + Z₂ + Z₀ + 3Z_f) |
| Line-to-line (LL) | Positive and negative in parallel | I_f = √3 E/(Z₁ + Z₂) |
| Double line-to-ground (LLG) | Negative ∥ zero, in series with positive | I_a1 = E/(Z₁ + Z₂Z₀/(Z₂ + Z₀)) |
For a solidly grounded generator with Z₀ small, the LG fault can be larger than the three-phase fault. Grounding the neutral through a reactance limits it.
Worked example
A generator has Z₁ = Z₂ = j0.2 pu and Z₀ = j0.1 pu, solidly grounded.
- Three-phase fault: 1/0.2 = 5 pu.
- LG fault: 3/(0.2 + 0.2 + 0.1) = 6 pu.
- LL fault: √3/0.4 = 4.33 pu.
Practice
A bus has a Thevenin reactance of 0.1 pu on a 100 MVA base. What is its fault level in MVA?
Show answer and explanation
Answer: 1000Fault MVA = S_base/X_th = 100/0.1 = 1000 MVA.
For the bus above at 11 kV, what is the three-phase fault current in kA? (Give 2 decimal places.)
Show answer and explanation
Answer: 52.49I = 1000 MVA/(√3 × 11 kV) = 52.49 kA.
The phase currents are I_a = 10, I_b = 0 and I_c = 0 A. What is the zero-sequence current in A? (Give 3 decimal places.)
Show answer and explanation
Answer: 3.333I₀ = (I_a + I_b + I_c)/3 = 10/3 = 3.333 A (and I₁ = I₂ = 3.333 A too).
Which fault involves all three sequence networks connected in series?
Show answer and explanation
Answer: AIn an LG fault, I_a1 = I_a2 = I_a0, so the three networks carry the same current in series.